Exercises — Semiconductor Electronics

6 questions✓ Free · step-by-step

11 markNCERT Exercises, Chapter 14
In an n-type silicon, which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants.
Hint. Decide first which dopant produces n-type material, then identify which carrier is in the minority.
An n-type semiconductor is made by doping silicon with a pentavalent element such as phosphorus, arsenic or antimony. Silicon has four valence electrons, so a pentavalent atom contributes one electron more than the four bonds require, and that spare electron is loosely bound and easily freed.

The result is that electrons are the majority carriers and holes are the minority carriers, the holes arising only from thermal generation.

Checking each option:

(a) Wrong dopant. Trivalent atoms produce p-type material, not n-type.

(b) Electrons are the majority carriers in n-type, not the minority.

(c) Correct on both counts: holes are indeed the minority carriers, and the dopant is indeed pentavalent.

(d) This describes p-type material.

Option (c) is a slightly indirect way of describing n-type silicon, since it names the minority carrier rather than the majority one, but every part of the statement is true.

✦ (c) Holes are minority carriers and pentavalent atoms are the dopants
21 markNCERT Exercises, Chapter 14
Which of the statements given in Exercise 14.1 is true for p-type semiconductors?
Hint. A p-type semiconductor is the mirror image of an n-type one; work out which option describes it directly.
A p-type semiconductor is made by doping silicon with a trivalent element such as boron, aluminium, indium or gallium. Having only three valence electrons, such an atom leaves one bond incomplete, and that vacancy behaves as a positively charged hole.

So in p-type material holes are the majority carriers and electrons are the minority carriers.

Looking back at the four options of Exercise 14.1, the statement that matches is (d): holes are majority carriers and trivalent atoms are the dopants.

Note that the material as a whole remains electrically neutral, because each dopant atom is itself neutral. Doping changes which carriers are available to move, not the total charge.

✦ (d) Holes are majority carriers and trivalent atoms are the dopants
31 markNCERT Exercises, Chapter 14
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to $(E_{g})_{C}$ , $(E_{g})_{S i}$ and $(E_{g})_{G e}$ . Which of the following statements is true? (a) $(E_{g})_{S i} < (E_{g})_{G e} < (E_{g})_{C}$ (b) $(E_{g})_{C} < (E_{g})_{G e} > (E_{g})_{S i}$ (c) $(E_{g})_{C} > (E_{g})_{S i} > (E_{g})_{G e}$ (d) $(E_{g})_{C} = (E_{g})_{S i} = (E_{g})_{G e}$
Hint. All three sit in the same group, so relate the band gap to atomic size as you move down the group.
All three elements belong to group 14 and have four valence electrons, but their band gaps differ widely:

$(E_{g})_{C} \approx 5.4 eV, (E_{g})_{S i} \approx 1.1 eV, (E_{g})_{G e} \approx 0.7 eV$

The trend follows from atomic size. Moving down the group the atoms get larger, so the valence electrons are further from the nucleus and less tightly held, and less energy is needed to lift one into the conduction band. The band gap therefore decreases down the group:

$(E_{g})_{C} > (E_{g})_{S i} > (E_{g})_{G e}$

This is why the three behave so differently. Carbon as diamond has so wide a gap that it is an insulator, while silicon and germanium have gaps small enough for thermal energy to promote some electrons, making them semiconductors.

✦ (c) (Eg)C > (Eg)Si > (Eg)Ge
41 markNCERT Exercises, Chapter 14
In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them. (b) they move across the junction by the potential difference. (c) hole concentration in p-region is more as compared to n-region. (d) All the above.
Hint. Ask what drives diffusion in general, and whether any potential difference exists before the junction has formed.
Diffusion is driven purely by a concentration gradient. Any species, charged or not, spreads from a region where it is abundant towards one where it is scarce, simply because random motion carries more particles out of the crowded region than into it.

In a p-n junction the p-side is rich in holes and the n-side is poor in them, so holes diffuse from p to n. Electrons diffuse the other way for the same reason.

Why the other options fail:

(a) Attraction by free electrons is not the mechanism; diffusion happens even for neutral particles, where no such attraction exists.

(b) This reverses cause and effect. The potential difference across the junction is created by the diffusion, as the departing carriers leave behind immobile charged ions. Before diffusion begins there is no barrier potential to move anything.

(d) Cannot be right, since (a) and (b) are both wrong.

The resulting barrier field eventually opposes further diffusion, and equilibrium is reached when diffusion and drift exactly balance.

✦ (c) hole concentration in p-region is more as compared to n-region
51 markNCERT Exercises, Chapter 14
When a forward bias is applied to a p-n junction, it (a) raises the potential barrier. (b) reduces the majority carrier current to zero. (c) lowers the potential barrier. (d) None of the above.
Hint. Compare the direction of the applied field with the direction of the junction's own barrier field.
In forward bias the positive terminal of the battery is connected to the p-side and the negative terminal to the n-side. The applied field therefore opposes the junction's own barrier field.

The net effect is that the potential barrier is lowered, the depletion region narrows, and majority carriers can cross the junction much more easily. The forward current therefore rises sharply once the applied voltage exceeds the knee, or cut-in, voltage — about 0.7 V for silicon and 0.3 V for germanium.

Checking the other options:

(a) This describes reverse bias, where the applied field reinforces the barrier field, widening the depletion region.

(b) The opposite happens: forward bias increases the majority carrier current rather than reducing it to zero.

(d) Incorrect, since (c) is right.

This asymmetry between forward and reverse bias is exactly what makes a diode useful as a rectifier.

✦ (c) lowers the potential barrier
62 marksNCERT Exercises, Chapter 14
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz? What is the output frequency of a full-wave rectifier for the same input frequency?
Hint. Count how many output pulses each type of rectifier produces per complete cycle of the input.
The answer follows from counting the pulses delivered per input cycle.

Half-wave rectifier. The diode conducts during only one half of each input cycle and blocks the other half. So each complete input cycle produces exactly one output pulse:

$f_{out} = f_{in} = 50 Hz$

Full-wave rectifier. Both halves of the input cycle are used, with the negative half inverted rather than discarded. Each input cycle therefore produces two output pulses:

$f_{out} = 2 f_{in} = 100 Hz$

The doubled frequency is a practical advantage as well as a numerical one, because the ripple in a full-wave output is both smaller and easier to smooth with a filter, since successive pulses arrive twice as often.

✦ Half-wave: 50 Hz. Full-wave: 100 Hz

Solutions written by the tuition.in editorial team and checked against leph206.pdf (NCERT, Reprint 2025-26 — the only version NCERT currently publishes for this chapter; the other 13 chapters are Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.

Take the next step